Under identical conditions, how many millilitres of 1 M- and 2 M- solutions are required to produce — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

To obtain the highest rise in temperature, the reactants must be in exact stoichiometric ratio (complete neutralization without any excess reactant to absorb heat). Let volume of $KOH$ be V ml. Then volume of $H_2SO_4$ is (100 - V) ml. Milliequivalents of $KOH$ = V * 1 = V. Milliequivalents of $H_2SO_4$ = (100 - V) * 2 * 2 = 4(100 - V). For complete neutralization: V = 4(100 - V) => 5V = 400 => V = 80 mL. Thus, 80 mL of 1 M $KOH$ and 20 mL of 2 M $H_2SO_4$ are required.